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Law of Floatation

The vapour de...

Question

The vapour density of a mixture containing $N O_{2}$ and $N_{2} O_{4}$ is $38.3$ at $33^{\circ} C$ calculate the no. of moles of $N O_{2}$ if $100 g$ of $N_{2} O_{4}$ were taken initially.

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Solution

Vapour density= $\frac{Effective molar mass}{2}$ -------(1)
$M_{e f f} = \frac{n_{1} M_{1} + n_{2} M_{2}}{n_{1} + n_{2}} = \frac{m_{1} + m_{2}}{n_{1} + n_{2}}$
$m_{1} + m_{2} = m = 100$ g
Initial moles of $N_{2} O_{4} = \frac{100}{92}$
From equation (1)
$3.83 = \frac{M_{e f f}}{2}$
$M_{e f f}$ =76.6
$N_{2} O_{4} ⇌ 2 N O_{2} m o l e s a t t = 0 \frac{100}{92} - a t t = t \frac{100}{92} - x 2 x$
Total moles= $\frac{100}{92} + x$
$M_{e f f} = \frac{n_{1} M_{1} + n_{2} M_{2}}{n_{1} + n_{2}} = \frac{m_{1} + m_{2}}{n_{1} + n_{2}} ⟹ 76.6 = \frac{(\frac{100}{92} - x) 92 + 2 x (46)}{\frac{100}{92} + x} ⟹ 76.6 = \frac{(100 - 92 x + 92 x) 92}{100 + 92 x} ⟹ 76.6 = \frac{9200}{100 + 92 x} ⟹ 100 + 92 x = \frac{9200}{76.6} ⟹ x = 0.218$
Thus, moles of $N O_{2}$ is $= 2 x = 2 \times$ $0.218 = 0.437$ moles

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